Some functions $\phi:\mathbb R\rightarrow \mathbb C$ that satisfy $\phi(x)\bar\phi(y)=\phi(x-y)$ Let us consider the function $\phi(x)$, $\phi:\mathbb R\rightarrow \mathbb C$ and indicate with $\bar \phi$ the conjugate of $\phi$. We consider the property: $$\phi(x)\overline{\phi(y)}=\phi(x-y)$$
The exponential function $e^{ix}$ ($x\in\mathbb R$), for example, satisfies this property. Is it the only function that satisfies this property? Do you know the other examples on this property?
Thank you very much.
 A: Assume $\overline{\phi(0)}\ne 0$, then $$\phi(0)=|\phi(0)|^2$$ so $\phi(0)>0$ therefore $\phi(0)=1$. From this we have $\phi(-x)=\overline{\phi(x)}$. As such, $1=\phi(x-x)=|\phi(x)|^2$, so really

$$\phi:\Bbb R\to S^1$$

and $\phi$ is a homomorphism by the given property. If you are willing to assume continuity, we know the continuous homomorphisms (i.e. group characters) from $\Bbb R\to S^1$ are just the maps
$$\{x\mapsto e^{i\alpha x}:\alpha\in\Bbb R\}\qquad (*)$$
So this is a complete characterization. In particular, since I added the little $\alpha$ in there, you have yourself uncountably infinitely many more examples!

You only asked about examples, but in case you're interested in how the classification of $(*)$ goes, it's not terribly hard if you know a little about basic topology (local compactness mostly) and it is presented very well in Tate's thesis or André Weil's Basic Number Theory (among other places).
A: Assuming that $\phi$ is not the zero function, we can show that if a function satisfies your functional equation then $\phi(0)=1$. This follows since $$\phi(x) = \phi(x+0) = \phi(x)\overline{\phi(0)}.$$
If we further assume that the function is differentiable at zero, then we can see that:
$$\phi'(x)=\lim_{h\to 0} \frac{\phi(x+h)-\phi(x)}{h} = \lim_{h\to 0} \frac{\phi(x)(\overline{\phi(-h)} - 1)}{h} = \phi(x) \lim_{h\to 0} \frac{\phi(h)-1}{h} = \phi(x) \phi'(0).$$
Here we used $\overline{\phi(-h)}=\phi(h)$ as was outlined in the comments. Thus $\phi$ satisfies the equation $\phi' = C\phi$ where $C = \phi'(0)$, and so we have $\phi = e^{C x}$ for some $C$.
To determine $C$, we re-examine the funcitonal equation. $e^{Cx}e^{\overline{C}y} = e^{C(x-y)}.$
Hence $Cx + \bar C y = C(x-y) + 2\pi\cdot k$ for some $k \in \mathbb{Z}$. This tells us that $\bar C y= - Cy + 2\pi \cdot k$. Further $\bar C + C = 2\pi \frac{k}y$. However, since $y$ can change, $k$ must be $0$. Thus $\bar C + C = 0$ and this means $C$ is imaginary.
